The result follows immediately from extended Ptolemy's theorem. THEOREM: If $ABCD$ is a cyclic quadrilateral where $AB = a , BC = b , CD = c , DA= d$ then, $AC^2 = (ac + bd)\frac{ad+bc}{ab+cd}$ and $BD^2 = (ac+bd) \frac{ab+cd}{ad+bc}$. Now, $AC > BD$ $\implies \frac{ad + bc}{ab + cd} > 1$ $\implies ad + bc - ab - cd >1$ $\implies (a-c)(d-b) > 0$ HENCE PROVED